Answer the following question.
An airplane travels 96 km with a tailwind in a duration that is 25% less than the duration it requires to fly the same distance against a headwind. If the airplane flies 36 km against the headwind in 1.5 hours, calculate the speed of the airplane in still air.

The correct option is 28 km/hr.

Step 1: Understand the given variables and setup
Let the speed of the airplane in still air be $v$ km/hr.
Let the speed of the wind be $w$ km/hr.

When flying with a tailwind (downstream/with the wind), the effective speed is:
$v_{\text{tailwind}}=v+w$

When flying against a headwind (upstream/against the wind), the effective speed is:
$v_{\text{headwind}}=v-w$

Step 2: Find the speed against the headwind
We are given that the airplane flies 36 km against the headwind in 1.5 hours.
Using the speed formula ($\text{Speed}=\frac{\text{Distance}}{\text{Time}}$):

$v_{\text{headwind}}=v-w=\frac{36\text{km}}{1.5\text{hours}}=24\text{km/hr}$

So, we have:
$v-w=24$ (Equation 1)

Step 3: Relate the durations for traveling 96 km
The time taken to fly 96 km against the headwind is:
$t_{\text{headwind}}=\frac{96}{v-w}=\frac{96}{24}=4\text{hours}$

The time taken to fly 96 km with a tailwind is 25% less than the duration against the headwind:
$t_{\text{tailwind}}=4\times (1-0.25)=4\times 0.75=3\text{hours}$

Step 4: Find the speed with the tailwind
Now we calculate the tailwind speed using the distance of 96 km and time of 3 hours:
$v_{\text{tailwind}}=v+w=\frac{96\text{km}}{3\text{hours}}=32\text{km/hr}$

So, we have:
$v+w=32$ (Equation 2)

Step 5: Solve for the speed of the airplane in still air ($v$)
Add Equation 1 and Equation 2:

$(v-w)+(v+w)=24+32$

$2v=56$

$v=\frac{56}{2}=28\text{km/hr}$

Thus, the speed of the airplane in still air is 28 km/hr.